Question

I'm trying to solve this problem

Calculate (a) how long it took King Kong to fall straight down form the top of the Empire State Building (380 m high), and (b) his velocity just before "landing".

My teacher said
try and rearagne this equation for t
X = Xo + Vo t + 2^-1 a t^2

I tried to rearange it for t and i guess i did it wrong i got

t = a^-1 (2(X - Xo - Vo))

I guess i did it wrong
so could you please show me how to rearange the equation for t step by step because I guess I did it wrong thanks...

Answer 1

If
X = vertical distance from Xo at time t, positive upwards, and
Vo = velocity at t=0
a = acceleration due to gravity, -9.81 m/s/s
Then the proposed equation is correct.
X = Xo + Vo t + 2^-1 a t^2

For your particular case,
Xo=380 m, (top of building)
X=0 (ground)
Vo=0 (just before the fall)
a=-9.81 m/s/s
That translates to

0 = 380 + 0*t + (-9.81)t²/2

Since the middle term "disappears" because of the coefficient of 0, then
9.81t²/2 = 380

from which you can find t without much ado.
The time is a touch more than 6 seconds if air resistance and his exceptional height are ignored.

Answer 2

thanks

Answer 3

Sorry, t should be a touch less than 9 seconds, being
sqrt(2*380/9.81)
The 6 seconds estimate did not account for the factor of 2.